Calculus

The position of a particle moving along the x-axis as a function of time,t, is given by x(t)=(1/6)t^3-t^2+3t-1 for t≥0. The particle's velocity becomes three times its initial velocity when t=?

I know v(t)=x'(t)=(1/2)t^2-2t+3=9, but I do not understand where does the 9 come from.

Some help would be appreciated, thanks.

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  1. read the question carefully.

    v(0) = 3

    you want v to be three times its initial value. 3*3 = 9

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